我知道如何使用密钥以传统方式删除节点 . 但是,我需要根据不同的变量(名称)删除节点,然后找到与该变量(名称)匹配的键,并删除包含该变量(名称)的所有节点 . 假设我的insert / inorder / FindMin函数和#include都出现在我的代码中 . 这是我的代码:

int totalStud;

struct evil {
    int key;
    char name[21];
    struct evil *right;
    struct evil *left;
};



struct evil *Delete(struct evil* root, int key){

    if (root == NULL){
        return root;
    }else if(key<root->key){
        root->left = Delete(root->left, key);
    }else if(key>root->key){
        root->right = Delete(root->right, key);
    }else{
        if (root->left == NULL){
            struct evil *temp = root->right;
            free(temp);
            return temp;
        }else if(root->right == NULL){
            struct evil *temp =  root->left;
            free(temp);
            return temp;
        }
        struct evil *temp = FindMin(root->right);
        root->key = temp->key;
        root->right = Delete(root->right,temp->key);

    }
}

struct evil *DelFinder(struct evil* root, char name[21]){ //I tried several things here, but I don't know what to do here to make it work.
    if (root == NULL){
        return root;
    }
    if(strcmp(root->name, name)==0){
        root = Delete(root, root->key);
        return root;
    }else{
        if (root->left != NULL){
            DelFinder(root->left,name);
        }
        if(root->right != NULL){
            DelFinder(root->right,name);
        }
    }
    return root;
}



int main(void){
    struct evil* root = NULL;

    totalStud = 5;
    root = Insert(root, 1015, "Jack");
    root = Insert(root, 1002, "Bob");
    root = Insert(root, 1085, "Sow");
    root = Insert(root, 1088, "Knack");
    root = Insert(root, 1099, "Sow");

    inorder(root); //all the elements in the BST before deleting Sow

    for (totalStud = 5; totalStud !=-1; totalStud--){
        root = DelFinder(root, "Sow");
    }
    inorder(root); //all the elemennts in BST after deleting Sow

}

我知道这是DelFinder函数的一个问题,但我不知道编写它的最佳方法是让它工作 . 我试图删除一个元素而不直接使用该键 . 如果我要求用户输入而不是在主函数中进行预设,那么阻止它们制作重复键的最佳方法是什么?因为要求用户放置一个不同的密钥,因为它已经存在? (重复的名字没问题) .